(D) Using the third equation of motion: $v^2 = u^2 + 2as$.Since the final velocity $v = 0$,we have $0 = u^2 - 2as$,which implies $s = \frac{u^2}{2a}$.

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(D) Using the third equation of motion: $v^2 = u^2 + 2as$.Since the final velocity $v = 0$,we have $0 = u^2 - 2as$,which implies $s = \frac{u^2}{2a}$.Assuming the deceleration $a$ is constant,the stopping distance $s$ is directly proportional to the square of the initial velocity: $s \propto u^2$.Therefore,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2$.Given $u_1 = 30 \, km/hr$,$s_1 = 8 \, m$,and $u_2 = 60 \, km/hr$:$\frac{s_2}{8} = \left( \frac{60}{30} \right)^2 = (2)^2 = 4$.$s_2 = 4 \times 8 = 32 \, m$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

$A$ car moving at a speed of $30 \, km/hr$ comes to a stop after traveling $8 \, m$ when the brakes are applied. If the same car is moving at a speed of $60 \, km/hr$,what distance (in $m$) will it travel before coming to a stop after the brakes are applied?

A
$8$
B
$16$
C
$24$
D
$32$

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Motion in Straight Line

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Uniformly Accelerated Motion

$A$ car moving with a velocity $6.25 \,ms^{-1}$ is decelerated with $2.5 \sqrt{v} \,ms^{-2}$ (where $v$ is the instantaneous velocity). The time taken by the car to come to rest is: (in $\,s$)

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$A$ train accelerates from rest at a constant rate $\alpha$ for distance $x_1$ and time $t_1$. After that,it retards to rest at a constant rate $\beta$ for distance $x_2$ and time $t_2$. Which of the following relations is correct?

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Which of the following decreases,in motion on a straight line,with constant retardation?

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$A$ particle travels $10\,m$ in the first $5\,s$ and $10\,m$ in the next $3\,s$. Assuming constant acceleration,what is the distance travelled in the next $2\,s$ (in $m$)?

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$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is

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$A$ car moving at a speed of $30 \, km/hr$ comes to a stop after traveling $8 \, m$ when the brakes are applied. If the same car is moving at a speed of $60 \, km/hr$,what distance (in $m$) will it travel before coming to a stop after the brakes are applied?

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$A$ car moving with a velocity $6.25 \,ms^{-1}$ is decelerated with $2.5 \sqrt{v} \,ms^{-2}$ (where $v$ is the instantaneous velocity). The time taken by the car to come to rest is: (in $\,s$)

$A$ train accelerates from rest at a constant rate $\alpha$ for distance $x_1$ and time $t_1$. After that,it retards to rest at a constant rate $\beta$ for distance $x_2$ and time $t_2$. Which of the following relations is correct?

Which of the following decreases,in motion on a straight line,with constant retardation?

$A$ particle travels $10\,m$ in the first $5\,s$ and $10\,m$ in the next $3\,s$. Assuming constant acceleration,what is the distance travelled in the next $2\,s$ (in $m$)?

$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is

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